w^2+41w=0

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Solution for w^2+41w=0 equation: 



w^2+41w=0

a = 1; b = 41; c = 0;
Δ = b2-4ac
Δ = 412-4·1·0
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$


$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-41}{2*1}=\frac{-82}{2}
=-41
$


$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+41}{2*1}=\frac{0}{2}
=0
$

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